Is $2\mathbb{Z}_{12}$ maximal ideal of $\mathbb{Z}_{12} \ ?$

Question

I came across this solution to the a/m problem here: https://sg.answers.yahoo.com/question/index?qid=20110515182700AAsSjeA.

The author of the solution claims that $\mathbb{Z}_{12}/2\mathbb{Z}_{12} \cong \mathbb{Z}_{6},$ hence $2\mathbb{Z}_{12}$ is not maximal ideal.

BUT $|\mathbb{Z}_{12}/2\mathbb{Z}_{12} |= 2$ , so shouldn't it be $\mathbb{Z}_{12}/2\mathbb{Z}_{12} \cong \mathbb{Z}_{2}$ and hence $2\mathbb{Z}_{12}$ is maximal ideal ?

Appreciate if anyone could advise on my doubts.

Answer 1

$2\mathbb Z_{12}$ is a subgroup under addition. Every element takes the form $2n+(12)$, where $n\in\mathbb Z$. Now, if $2k+(12)=2\ell+(12)$, then $2(k-\ell)\in(12)$, so $6\mid k-\ell$. Thus, if you pick $n$ from $\{0,1,2,3,4,5\}$, you get distinct elements of $2\mathbb Z_{12}$. That is, $|2\mathbb Z_{12}|\geq 6$. By Lagrange's Theorem, it is an equality. Therefore $|\mathbb Z_{12}/2\mathbb Z_{12}|=12/6=2$ as you observed.

Answer 2

Yes, $2\mathbb{Z}_{12}$ is a maximal ideal because $$\mathbb{Z}_{12}/2\mathbb{Z}_{12} \cong \mathbb{Z}_{2}$$ which is a field